At a particular temperature, 4.00 mole of NOCI is placed in a 1.00 L rigid container, and decomposes according to the reaction below. At equilibrium, 3.00 mol of NOCI is present. Calculate the equilibrium constant Kc for this reaction.
2NOCI(g) = 2NO(g)+Cl2(g)
Chemistry- How would I go about solving for Kc when I don't have K?
2020-10-24 11:54 pm
回答 (1)
2020-10-25 12:59 am
✔ 最佳答案
2NOCI(g) ⇌ 2NO(g) + Cl₂(g) KcInitial (mol/L): 4.00 0.00 0.00
Change (mol/L): -2y +2y +y
Eqm (mol/L): (4.00 - 2y) 2y y
At equilibrium:
[NOCl] = (4.00 - 2y) mol/L = 3.00 mol/L, Hence, y = 0.500 mol/L
[NO] = 2y mol/L = 2 × 0.500 mol/L = 1.00 mol/L
[Cl₂] = 0.500 mol/L
Kc = [NO]² [Cl₂] / [NOCl]² = (1.00)² (0.500) / (3.00)² = 0.0556
收錄日期: 2021-04-23 23:07:36
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